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C h a p t e r 1 I n t ro d u c t io n t o L i n e a r P r o g r a m m i n g

15. Consider the l inear programming problem

M aximize z = cTx

subject to

A x < b

x > _ O .

If x~ and x z are feasible solutions, show that

1 2

X ~ X 1

I

~ X 2

is a feasible solution.

16. Generalize the previous exercise to show that

x - -- - rx 1 - I - SX 2

is a feasible solution if r + s = 1.

1 3 GEOMETRY OF L INEAR PROGRAM MING PROBLEMS

I n t hi s s e c ti o n w e c o n s i d e r t h e g e o m e t r y o f l i n e a r p r o g r a m m i n g p r o b -

l e m s b y f ir st l o o k i n g a t th e g e o m e t r i c i n t e r p r e t a t i o n o f a si n gl e c o n s t r a i n t ,

t h e n a t a s e t o f c o n s t r a i n t s , a n d f i n a l ly a t t h e o b j e c t i v e fu n c t i o n . T h e s e

i d e a s g i v e r i s e t o a g e o m e t r i c m e t h o d f o r s o l v i n g a l i n e a r p r o g r a m m i n g

p r o b l e m t h a t i s s u c c e ss f u l o n l y f o r p r o b l e m s w i th t w o o r t h r e e v a r i a b le s .

H o w e v e r , t h e g e o m e t r i c c o n c e p t s t h a t w e d i s c u s s c a n b e b u i l t i n t o a n

a l g e b r a i c a l g o r i t h m t h a t c a n e f f e c ti v e ly s o lv e v e r y la r g e p r o b l e m s . A f t e r

c a s t i n g t h e g e o m e t r i c i d e a s i n a l i n e a r a l g e b r a s e t t i n g , w e w i l l p r e s e n t t h i s

a l g o r i t h m m t h e s im p l ex m e t h o d m i n S e c t io n 2 .1 .

Geometry of a Constraint

A s in g le c o n s t r a in t o f a l i ne a r p r o g r a m m i n g p r o b l e m i n s t a n d a r d f o r m ,

s a y t h e i t h o n e ,

a i l x I -t- a i 2 x 2 -t- " " + a i n X n < b i ,

c a n b e w r i t t e n a s

w h e r e

aTx < b i ,

a T - - [ a i l

a i 2 . . . a i n ] .

T h e s e t o f p o i n t s x = ( x 1, x 2 , . . . , x , ) i n R ~ th a t s a ti s fy t h i s c o n s t r a i n t is

c a l l e d a c lo s e d h a l f - s p a c e . I f t h e i n e q u a l i t y i s r e v e r s e d , t h e s e t o f p o i n t s

x = ( X l, x 2 , . . . ,

X n )

in R n sa t i s fy ing

aTx > b i

i s a l so ca l l ed a c losed ha l f - space .


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1.3 Geometry of Linear Programming Problems

71

EXAMPLE 1.

sp a c e

C o n s i d e r t h e c o n s t r a i n t 2 x + 3 y < 6 a n d t h e c l o s e d h a lf -

- [y] 3 [y] 6 /

w h i c h c o n s i st s o f t h e p o i n t s s a t i sf y in g t h e c o n s t r a i n t. N o t e t h a t t h e p o i n t s

( 3, 0 ) a n d ( 1, 1 ) s a t is f y t h e i n e q u a l i t y a n d t h e r e f o r e a r e i n H . A l so , t h e

p o i n t s (3 , 4 ) a n d ( - 1 , 3 ) d o n o t s a t is fy t h e i n e q u a l i t y a n d t h e r e f o r e a r e n o t

in H . Ev e r y p o in t o n t h e l i n e 2x + 3y = 6 s a t is f ie s th e c o n s t r a in t a n d t h u s

l ies in H . A

W e c a n g r a p h a c l o se d h a l f - s p a c e i n R E b y g r a p h i n g t h e l in e a n d t h e n

u s in g a t e s t p o in t t o d e c id e w h ic h s id e o f t h e l i n e i s i n c lu d e d in t h e

h a l f - sp a c e . A s imp le w a y to g r a p h th e l i n e i s t o f i n d i t s x - a n d y - in t e r c e p t s .

B y se t t i n g y - 0 i n t h e e q u a t io n o f t h e l i n e a n d so lv in g f o r x , w e o b t a in

th e x - in t e r c e p t a n d p lo t i t o n t h e x - a x i s . S imi l a r ly , b y s e t t i n g x = 0 i n t h e

e q u a t i o n o f t h e l i n e a n d s o l v in g f o r y , w e o b t a i n t h e y - i n t e r c e p t a n d p l o t it

o n t h e y - ax is . W e n o w c o n n e c t t h e t w o p o i n t s t o s k e t c h t h e g r a p h o f t h e

l i n e . T o c h o o se a t e s t p o in t , w e c h e c k w h e th e r t h e o r ig in i s o n t h e l i n e . I f i t

i s n o t , w e u se i t a s a t e s t p o in t , c h e c k in g w h e th e r t h e o r ig in s a t i s f i e s t h e

in e q u a l i t y . I f i t d o e s , t h e s i d e o f t h e l i n e c o n t a in in g t h e o r ig in ( o u r t e s t

p o in t ) i s t h e c lo se d h a l f - sp a c e H . I f i t d o e s n o t , t h e n t h e o th e r s i d e o f t h e

l i n e i s t h e c lo se d h a l f - sp a c e . I f t h e o r ig in i s o n t h e l i n e , so me o th e r p o in t

n o t o n t h e l i n e m u s t b e s e l e c t e d a s t h e t e s t p o i n t . S o m e p o s s i b l e c h o i c e s

are (1, 0), (0, 1) , or (1, 1) .

EX AMPLE 1 ( c o n t in u e d ) . W e c o m p u te t h e x - in t e r c e p t t o b e x = 3

a n d t h e y - i n t e r c e p t t o b e y = 2 . T h e s e p o i n t s h a v e b e e n p l o t t e d a n d t h e

l i n e c o n n e c t i n g t h e m h a s b e e n d r a w n i n F i g u r e 1 . 2 a . S i n c e t h e o r i g i n d o e s

n o t l i e o n t h e l i n e 2x + 3y = 6 , w e u se t h e o r ig in a s t h e t e s t p o in t . T h e

c o o r d in a t e s o f t h e o r ig in sa t is f y t h e i n e q u a l i t y , so t h a t H l ie s b e lo w th e

l i ne a n d c o n ta in s t h e o r ig in a s sh o w n in F ig u r e 1 .2b . A

FIGURE 1 2

Closedhalf-space in two dime nsions.


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Chapter 1 Introduction to Linear Programming

T

z

~ 0 , 0 , 4 ) 2 0

(5,0,

FIGURE 1 3 Closedhalf-space in th ree dim ensions.

E X AMPLE 2 . Th e c o n s t r a i n t i n t h r e e v a r i a b l e s , 4 x + 2 y + 5 z < 2 0 ,

d e f i n e s t h e c l o s e d h a l f - s p a c e H i n R 3, w h e r e

H = [4 2 5 ] < 20 .

W e c a n g r a p h H i n R 3 b y g r a p h i n g t h e p l a n e 4 x + 2 y + 5 z = 2 0 a n d

c h e c k i n g a te s t p o i n t. T o g r a p h t h e p l a n e , w e g r a p h t h e i n t e r s e c t i o n o f t h e

p l a n e w i t h e a c h o f t h e c o o r d i n a t e p l a n e s . T h e s e i n t e r s e c t i o n s a r e l i ne s in

t h e c o o rd i n a t e p l a n e s . T h u s , l e t t i n g z = 0 y i e l d s t h e l i n e 4 x + 2 y = 2 0 i n

t h e x y p l a n e , w h i c h c a n b e g r a p h e d a s d e s c r i b e d i n E x a m p l e 1 . S i m i l a r l y ,

l e t t ing y = 0 y ie lds the l ine 4x + 5z = 20 in the x z p l a n e . F i n a l l y , s e t t i n g

x = 0 y i e l d s t h e l i n e 2 y + 5 z = 2 0 i n t h e y z p l a n e . T h e g r a p h o f th e p l a n e

c o n t a i n i n g t h e s e l i n e s i s s h o w n i n F i g u re 1 . 3 . Th e o r i g i n d o e s n o t l i e o n t h e

p l a n e a n d t h u s c a n b e u s e d a s a t e s t p o i n t . I t s a t i s f i e s t h e i n e q u a l i t y s o t h a t

t h e c l o s e d h a l f - s p a c e c o n t a i n s t h e o r i g i n a s s h o w n i n F i g u re 1 .3 . A

I n m o r e t h a n t h r e e d i m e n s i o n s , i t i s i m p o s s i b l e t o s k e t c h a c l o s e d

h a lf - sp a c e . H o w e v e r , w e c a n t h i n k a b o u t t h e g e o m e t r y o f c lo s e d h a lf - s pa c e s

i n a n y d i m e n s i o n a n d u s e t h e l o w e r d i m e n s i o n e x a m p l e s a s m o d e l s f o r o u r

c o m p u t a t i o n s .

A t y p ic a l c o n s t r a in t o f a li n e a r p r o g r a m m i n g p r o b l e m i n c a n o n i c a l f o r m

h a s t h e e q u a t i o n

aTx = b . (1 )


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1.3 Geom etry of L inear Program ming Problems ~

I t s g raph i n

R n

is a hyperp l ane . I f t h i s equa t i o n w ere an i nequa l i t y ,

nam el y ,

aTx _< b,

t hen t he s e t o f po i n t s s a t i s fy i ng t he i nequa l i t y wou l d be a c l o sed ha l f - space .

Thus , a hyperp l ane i s t he boundary o f a c l o sed ha l f - space . In t u i t i ve l y , i t

cons i st s o f t he po i n t s t ha t a re i n the ha l f - space , bu t on i t s edge .

EXAM PLE 3 . Th e equ a t i on 4x + 2y + 5z = 20 de f i nes a hyp erp l a ne

in R 3. Th e gra ph of th is hy per p lan e , w hich i s rea l ly a p la ne in th i s case , is

shown in F i gu re 1 .3 . Th e hyp erp l a ne H is t he bo un dar y o f t he c l o sed

ha l f - space H 1 de f i ned by t he i neq ua l i t y 4x + 2y + 5z < 20 , con s i de re d i n

^

E xam pl e 2 . Th e ha l f - space H 1 ex t ends be l ow t he h ype rp l an e H and l ie s

b e h i n d t h e p a g e . W e a l s o s e e t h a t H is t h e b o u n d a r y o f t h e c l o s e d

ha l f - space H 2 de f i ned by t he i nequa l i t y 4x + 2y + 5z > 20 . Th e ha l f - space

H 2 e x t e n d s a b o v e t h e h y p e r p l a n e a n d r e a c h e s o u t o f t h e p a g e . /x

The hyp erp l a ne H de f i ned by (1 ) d iv i des R ~ i n t o t he t wo c l o sed

hal f - spaces

H i = {x ~ R " l a T _b} .

W e a l so s ee t ha t H 1 n H 2 = H , t he o r i g i na l hyperp l an e . In o t h e r w ords , a

hyperp l ane i s t he i n t e r s ec t i on o f t wo c l o sed ha l f - spaces .

R eca l l f rom S ec t i on 1 .2 t ha t a f eas i b le so l u t i on t o a l i nea r p ro g ra m m i ng

prob l em is a po i n t i n R n t ha t s a ti sf ie s a l l t he cons t r a i n t s o f t he p rob l em . I t

t hen fo l lows t ha t t h i s s e t o f f eas i b le so l u t i ons i s t he i n t e r s ec t i on o f a ll t he

c l o sed ha l f - spaces de t e rm i ned by t he cons t r a i n t s . S pec i f i ca l l y , t he s e t o f

so lu t ions to an inequ al i ty ( < or > ) co ns t ra in t i s a s ing le c losed hal f - spac e ,

whereas t he s e t o f so l u t i ons t o an equa l i t y cons t r a i n t i s t he i n t e r s ec t i on o f

two c losed hal f - spaces .

Sketch the se t o f a l l feas ib le so lu t ions sa t i s fy ing the se t o f

XAM PLE 4.

i nequa l i t i e s

2x + 3y < 6

- x + 2 y < 4

x > 0

y > 0 .

Solution.

Th e se t o f so lu t ions to th e f i rs t inequa l i ty , 2 x + 3y _< 6 , is

shown as t he shaded r eg i on i n F i gu re 1 .4a and t he s e t o f so l u t i ons t o t he


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~ Chapter 1 Introduction to Linear Programming

second inequa l i ty , - x + 2y < 4, f o r m the shaded r eg ion in F igur e 1 .4b . I n

de t e r min ing the se r eg ions , we have used the o r ig in a s a t e s t po in t . The

r eg ions sa t i s f y ing the t h i r d and f our th cons t r a in t s a r e shown in F igur e s

1.4c and 1 .4d, r espec t ive ly . T he po int (1 , 1) was u sed as a tes t po int to

de t e r m ine the se r eg ions . Th e in t e r sec t ion o f t he r eg ions in F igur e s

1.4a- l .4d i s shown in Figure 1 .4e ; i t i s the se t of a l l f eas ib le solut ions to

the g iven se t of cons t ra in ts . /x


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1.3 Geometry of Linear Programming Problems 75

3

)"

FIGURE 1.5 Set of all feasible solut ions (thr ee dim ension s).

EXAM PLE 5 . Us i n g the s am e t ech n i qu e a s in t he p rev i ou s exam pl e , we

f i nd t he ske t ch o f t he r eg i on i n R 3 def i ned by t he i nequa l i t i e s

x > 0

y > 0

z > 0

5 x + 3 y + 5 z < 1 5

10x + 4y + 5z < 20 .

The f i r s t th ree inequal i t i es l imi t us to the f i r s t oc tan t o f

R 3.

T h e o t h e r t w o

i nequa l i t i e s de f i ne ce r t a i n c l o sed ha l f - spaces . The r eg i on , wh i ch i s

t he i n t e r s ec t i on o f t hese t wo ha l f - spaces i n the f ir st oc t an t , i s show n i n

Figure 1.5.

Geom etry of the Ob jective Function

T h e o b j e c t i v e f u n c t i o n o f a n y l i n e a r p r o g r a m m i n g p r o b l e m c a n b e

wr i t t en a s

c T x .

I f k i s a cons t an t , t hen t he g raph o f t he e qua t i on

cXx = k

is a h y p e r p la n e . A s s u m e t h a t w e h a v e a l i n e a r p r o g r a m m i n g p r o b l e m t h a t

a sks fo r a m ax i m um va l ue o f t he ob j ec t i ve func t i on . In so l v i ng t h i s

p rob l em , we a re s ea rch i ng fo r po i n t s x in t he s e t o f f eas i b le so l u t i ons fo r

wh i ch t he va l ue o f k i s a s l a rge a s pos s i b l e . Geom et r i ca l l y we a re l ook i ng


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7 ~ C h a p t e r 1 I n t ro d u c t io n t o L i n e a r P r o g ra m m i n g

f o r a h y p e r p l a n e t h a t i n t e r s e c t s t h e s e t o f f e a s ib l e s o l u t io n s a n d f o r w h i c h

k i s a m a x i m u m . T h e v a l u e o f k m e a s u r e s t h e d i s t a n c e f r o m t h e o r i g i n t o

t h e h y p e r p l a n e . W e c a n t h i n k o f s t a rt i n g w i t h v e ry l a r g e v a l u e s o f k a n d

t h e n d e c r e a s i n g t h e m u n t i l w e f i n d a h y p e r p l a n e t h a t j u s t t o u c h e s t h e s e t

o f f e a s ib le so lu t ions .

EX AM PLE 6 . C o n s i d e r t h e l i n e a r p r o g r a m m i n g p r o b l e m

M a x i m i z e z = 4 x + 3 y

sub je c t t o

x + y < 4

5x + 3y < 15

x > 0 , y > 0

T h e s e t o f f e a si b l e s o l u t io n s ( t h e s h a d e d r e g i o n ) a n d t h e h y p e r p l a n e s

z = 9 , z = 12, z = ~ , a n d z = 15

a r e s h o w n i n F i g u r e 1 . 6 . N o t e t h a t i t a p p e a r s t h a t t h e m a x i m u m v a l u e o f

5

t he ob je c t ive func t ion i s -~ , wh ic h i s ob ta in e d w he n x = 3 , y = 3 - Th i s

c on je c tu r e wi ll be ve r i f ie d in a l a t e r se c t ion . A

A l i n e a r p r o g r a m m i n g p r o b l e m m a y n o t h a v e a s o l u t io n i f t h e s e t o f

f e a s i b l e s o l u t i o n s i s u n b o u n d e d . I n t h e f o l l o w i n g e x a m p l e , w e a r e a s k e d t o

ma x im iz e the va lue o f the ob je c t ive func t ion , bu t we d i sc ove r tha t no suc h

ma ximum e x i s t s .

FIfURE 1 6

Objectivefunction hyperplanes (two dimensions).


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1.3 Geometryof Linear Programming Problems ~

EX AM PLE 7. C o n s i d e r th e l i n e a r p r o g r a m m i n g p r o b l e m

M a x i m i z e z = 2 x + 5 y

sub je c t t o

- 3 x + 2 y < 6

x + 2 y > 2

x > 0 , y > 0

T h e g r a p h o f t h e s e t o f f e a s ib l e s o l u t i o n s i s s h o w n a s th e s h a d e d r e g i o n i n

F i g u r e 1 .7. W e h a v e a ls o d r a w n t h e g r a p h s o f t h e h y p e r p l a n e s

z = 6 , z = 14, a nd z = 2 0 .

W e s e e t h a t i n e a c h c a s e t h e r e a r e p o i n t s t h a t l ie t o t h e r i g h t o f t h e

hy pe rp la ne a nd t ha t a r e s ti ll in the se t o f f e a s ib le so lu t ions . Ev ide n t ly the

v a l u e o f t h e o b j e c ti v e f u n c t i o n c a n b e m a d e a r b i t ra r i l y l a rg e . A

EX AM PLE 8. C o n s i d e r a li n e a r p r o g r a m m i n g p r o b l e m t h a t h a s t h e

sa m e se t o f c ons t r a in t s a s in Ex a m ple 7 . Ho we ve r , a s s um e th a t i t is a

m i n i m i z a t i o n p r o b l e m w i t h o b j e c t i v e f u n c t i o n

z = 3x + 5y.

W e h a v e d r a w n t h e g r a p h o f th e s e t o f f e a s ib l e s o l u t i o n s i n F i g u r e 1 .8 ( t h e

s h a d e d r e g i o n ) a n d t h e h y p e r p l a n e s

z = 6 , z = 9 , a nd z = 15.

I t a p p e a r s t h a t t h e o p t i m u m v a l u e o f t h e o b j e c t iv e f u n c t i o n i s z = 5 w h i c h

i s ob ta ine d whe n x = 0 , y = 1 . Sma l l e r va lue s o f the ob je c t ive func t ion ,

s u c h a s z = 3 , y i e ld g r a p h s o f h y p e r p l a n e s t h a t d o n o t i n t e r s e c t t h e s e t o f


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78

C h a p t e r 1 I n t r o d u c t io n to L i n e a r P r o g r a m m i n g

FIGURE 1 8

Minimization two dimensions).

f e a si b le s o l u ti o n s. Y o u m a y w a n t t o s k et c h t h e h y p e r p l a n e s c o r r e s p o n d i n g

to z = 3 an d z = 5 . A

Geom etry of the Set of Feas ible Solutions

W e n o w e x p l o r e t h e q u e s t i o n o f w h e r e i n t h e s e t o f f e a s i b le s o l u t i o n s

w e a r e l i k e l y t o f in d a p o i n t a t w h i c h t h e o b j e c t i v e f u n c t i o n t a k e s o n i t s

o p t i m a l v a l u e . W e f i rs t s h o w t h a t i f x 1 a n d x 2 a r e t w o f e a s i b l e s o l u t i o n s ,

t h e n a n y p o i n t o n t h e l in e s e g m e n t j o i n i n g t h e s e t w o p o i n t s i s a l so a

f e a s i b l e s o l u t i o n . T h e l i n e s e g m e n t j o i n i n g x I a n d x 2 i s d e f i n e d a s

{x ~ R n Ix

= ,~x I q-

(1

- / ~ ) x 2 ,

0 __< A __< 1}.

O b s e r v e t h a t , i f A = 0 , w e g e t x 2 a n d , i f A = 1 , w e g e t x 1. T h e p o i n t s o f

t h e l i n e s e g m e n t a t w h i c h 0 < A < 1 a r e c a l l e d t h e i n t e r i o r p o i n t s o f t h e

l i n e s e g m e n t , a n d x 1 a n d x 2 a n d c a l l e d i ts

end points.

N o w s u p p o s e t h a t x 1 a n d x 2 a r e f e a s i b le s o l u t i o n s o f a l i n e a r p r o g r a m -

m i n g p r o b l e m . I f

aTx < b i

is a c o n s t r a i n t o f t h e p r o b l e m , t h e n w e h a v e

8TXl _~ b i a n d aTx2 __< b i .


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1 . 3 G e o m e t r y o f L i n e a r P r o g r a m m i n g P r o b l e m s ~

F o r a n y p o i n t x = Ax 1 + (1 - ) k ) x 2 , 0 _~ < , ~ _ ~ < 1 , o n t h e l in e s e g m e n t j o i n i n g

X 1 a n d x 2 , w e h a v e

a T x = a T ( A X l +

(1

- A ) x 2 )

= AaTXl +

(1

- A ) a T x 2


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8 0 Chapter 1 Introduction to Linear Programming

S i s c o n v e x if , w h e n e v e r x I a n d x 2 ~ S , s o d o e s

x - -

A X 1 +

(1

- A ) x 2

fo r 0 < A < 1 . A

E X AM P LE 9 . T h e s e t s i n R E i n F i g u r e s 1 .9 a n d 1 . 1 0 a r e c o n v e x . T h e

s e t s i n R E i n F i g u r e 1 .1 1 a r e n o t c o n v e x . / x

T h e f o l l o w i n g r e s u l t s h e l p t o i d e n t i f y c o n v e x s e t s .

THEO REM 1 .1 .

A c l o s e d h a l f - s p a c e i s a c o n v e x s e t .

P r o o f .

L e t t h e h a l f - s p a c e H 1 b e d e f i n e d b y c T x _< k . L e t x 1 a n d

X 2 ~ H i a n d c o n s i d e r x = A x 1 + (1 - A ) x 2 , ( 0 < A < 1).

T h e n

c T x = c T [ , ~ X 1 "J- (1 - , ~ ) x 2 ]

- - A c T x 1 + (1 - A ) c T x 2 .


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1 .3 Geom etry o f L inear Programm ing Problems 81

Si nc e h >_ 0 a n d 1 - h >_ O, w e o b ta in

T h u s ,

c T x < A k +

( 1 - A ) k = k .

c T x

_< k,

so th a t x ~ H 1. A

THEOREM 1.2 .

A h y p e r p l a n e i s a c o n v e x s e t.

P r o o f E x e r c i s e . A

THEOREM 1.3 . T h e i n t e r se c t i o n o f a f i n i t e c o l l e c t io n o f c o n v e x s e t s is

c o n o e x .

P r o o f . E x e r c i s e . A

THEOREM 1.4 . L e t A b e a n m n m a t r i x , a n d l e t b b e a ve c t o r i n R m .

T h e s e t o f s o l u t i o n s t o t h e s y st e m o f l i n e a r e q u a t i o n s A x = b , i f i t i s n o t e m p t y ,

i s a c o n v e x s e t .

P r o o f . E x e r c i s e . A

C o n v e x s e ts a r e o f t w o t yp e s: b o u n d e d a n d u n b o u n d e d . T o d e f i n e a

b o u n d e d c o n v e x s e t, w e f ir st n e e d t h e c o n c e p t o f a r e c t a n g le . A r e c t a n g l e

in

R ~ i s a s e t ,

R = { x ~ R n l a i 0 3 x + y < 6

x > _ O , y > _ O

3. 4.

4 x + y > 8 3 x + y < 6

3x + 2y > 6 2x + 3y > 4

x > 0 , y > 0 x > 0 , y > 0


13/32

8~ Chap t e r 1 In t roduc t ion t o L i ne ar Programm i ng

2 x + 5 y + 5 z < 2 0 4 x + 5 y + 4 z < 2 0

4 x + 2 y + z < 8 2 0 x + 1 2 y + 1 5 z < 6 0

x > 0 , y > 0 , z > 0 x > 0 , y > 0 , z > 0

In Exerc i se s 7 -12 ske t ch t he se t o f f eas i b l e so l u t i ons t o t he g i ven se t o f i nequa l i -

ties.

11.

7. 8.

- x + y < 2 x + y < 3

2 x + y < 4 2 x + y < 4

x > 0 , y > 0 x > 0 , y > 0

10.

x + y > 3 - x + y < 2

- 3 x + 2 y < 6 2 x + y < 2

x > 0 , y > 0 y < l

x > 0

12.

6 x + 4 y + 9 z < 3 6 1 2 x + 6 y + 1 6 z < 8 4

2 x + 5 y + 4 z < 2 0 8 x + 5 y + 1 2 z < 6 0

x > 0 , y > 0 , z > 0 x > 0 , y > 0 , z > 0

In Exerc i se s 13- 16 ( a ) ske t ch t he se t o f f eas i b l e so l u t ions t o t he g i ven l i nea r

p ro gram mi n g p rob l em , (b ) d r aw t he ob j ec t i ve func t i on z = cTx = k , f o r t he i nd i-

ca t ed va l ues o f k , and ( c ) con j ec t u re t he op t i ma l va l ue o f z .

13. Maximize z = 3x + 4y

sub j ec t t o

k = 6, 8, 10, and 12.

14. Maximize z = 2x + 3y

subjec t to

k = 4, 6, 8, and 10.

15. Maximize z = 3x + y

sub j ec t t o

k -- 2, 6, 8, and 12.

x + 3 y < 6

4x + 3y < 12

x > 0 , y > 0

x + y < 4

3 x + y < 6

x + 3 y < 6

x > 0 , y > 0

- 2 x 3 y < 6

x + y < 4

3 x + y < 6

x > 0 , y > 0


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1.3 Geom etry o f Linea r Programm ing Problems 8~

1 6 . M a x i m i z e z = 4 x 1 + 8 x 2 + x 3

s u b j e c t t o

8 X 1 -~-

2 x 2 + 5 x 3 < 6 8

5 x 1 + 9 x z + 7 x 3 < 1 2 0

1 3 x 1 + l l x 2 + 4 3 x 3 < 2 5 0

x 1 > 0 , x 2 > 0 , x 3 > 0

k = 8 0 , 9 0 , 1 0 0 , an d 1 1 0 .

I n E x e r c i s e s 1 7 - 2 4 d e t e r m i n e w h e t h e r t h e g i v e n s e t is c o n v e x.


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84 C h a p t e r 1 I n t r o d u c t io n t o L i n e a r P r o g r a m m i n g

2 5 . P r o v e t h a t R " i s a convex se t .

2 6 . P r o v e t h a t a s u b s p a c e o f

R n

i s a convex se t .

2 7 . S h o w t h a t a r e c t a n g l e i n R " i s a c o n v e x se t .

2 8 . L e t H 2 b e t h e h a l f - s p a c e i n R " d e f i n e d b y cX x > k . S h o w t h a t H 2 i s c o n v e x .

2 9. S h o w t h a t a h y p e r p l a n e H i n R " i s c o n v e x ( T h e o r e m 1 .2 ).

3 0. S h o w t h a t t h e i n t e r s e c t i o n o f a fi n i te c o l l e c t i o n o f c o n v e x s e ts i s c o n v e x

( T h e o r e m 1 . 3 ) .

3 1. G i v e t w o p r o o f s o f T h e o r e m 1 .4 . O n e p r o o f s h o u l d u s e t h e d e f i n i t i o n o f c o n v e x

s e t s a n d t h e o t h e r s h o u l d u s e T h e o r e m 1 . 3 .

3 2. C o n s i d e r t h e l i n e a r p r o g r a m m i n g p r o b l e m

M a x i m i z e z - - cTx

s u b j e c t t o

A x _ < b

x > _ O .

L e t x ~ a n d x 2 b e f e a s i b l e s o l u t i o n s t o t h e p r o b l e m . S h o w t h a t , i f t h e o b j e c t i v e

f u n c t i o n h a s t h e v a l u e k a t b o t h x 1 a n d x 2 , t h e n i t h a s t h e v a l u e k a t a n y p o i n t

o n t h e l i n e s e g m e n t j o i n i n g x I a n d x 2 .

3 3. S h o w t h a t t h e s e t o f a ll s o l u t i o n s t o A x < b , i f i t i s n o n e m p t y , i s a c o n v e x s e t .

3 4. S h o w t h a t t h e s e t o f s o l u t i o n s t o A x > b , i f i t i s n o n e m p t y , i s a c o n v e x s e t .

3 5 . A f u n c t i o n m a p p i n g R " i n t o

R m

is c a l l ed a l i n e a r t r a n s f o r m a t i o n i f f ( u + v )

= f ( u ) + f ( v ) , f o r a n y u a n d v i n R n, a n d f ( r u ) = r f ( u ), fo r a n y u i n R " a n d r

i n R . P r o v e t h a t i f S i s a c o n v e x s e t i n R n a n d f i s a l i n e a r t r a n s f o r m a t i o n

m a p p i n g R " i n t o

R m,

t h e n

f ( S )

= { f ( v ) I v ~ S } i s a c o n v e x s e t. A f u n c t i o n f

d e f i n e d o n a c o n v e x s e t S i n R " i s c a l l e d a c o n v e x f u n c t i o n i f

f ( A x 1 + (1 - - A)x 2 ) < Af (x 1 ) + (1 - - A ) f ( x 2 )

f o r 0 _ < A _ < l a n d a n y x ~ , x 2 ~ S .

3 6 . S h o w t h a t a f u n c t i o n f d e f i n e d o n a c o n v e x s e t S i n R " i s c o n v e x i f t h e l i n e

s e g m e n t j o i n i n g a n y t w o p o i n t s ( X l, f ( x l ) ) a n d ( x e , f ( x e ) ) d o e s n o t l ie b e l o w i ts

g r a p h . ( S e e F i g u r e 1 . 1 2 . )

3 7. S h o w t h a t t h e o b j e c t i v e f u n c t i o n z = c X x o f a l i n e a r p r o g r a m m i n g p r o b l e m i s a

c o n v e x f u n c t i o n .

f ( x 2 )

f ( x l )

t I

x 1 x2

F I G U R E 1 1 2


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1 .4 T h e E x t r e m e P o i n t T h e o r e m 85

1 4 THE EXTREM E POINT THEOREM

W e c o n t i n u e i n t h is s e c ti o n t o w a r d o u r g o a l o f u n d e r s t a n d i n g t h e

g e o m e t r y o f a l i n e a r p r o g r a m m i n g p r o b l e m . W e f i r s t c o m b i n e t h e r e s u l t s

o f t h e l a st s e c t io n t o d e s c r i b e t h e g e o m e t r y o f a g e n e r a l l i n e a r p r o g r a m -

m i n g p r o b l e m a n d t h e n i n t r o d u c e t h e c o n c e p t o f a n e x t r e m e p o i n t, o r

c o m e r p o i n t . T h e s e b e c o m e c a n d i d a t e s f o r s o l u t i o n s t o t h e p r o b l e m .

W e n o w c o n s i d e r a g e n e r a l l i n e a r p r o g r a m m i n g p r o b l e m . T h e g r a p h o f

e a c h c o n s t r a i n t d e f i n e d b y a n i n e q u a l i t y i s a c l o s e d h a l f - s p a c e . T h e g r a p h

o f e a c h c o n s t r a i n t d e f i n e d b y a n e q u a l i t y is a h y p e r p l a n e , o r i n t e r s e c t i o n o f

t w o c l o s e d h a l f - s p a c e s . Th u s , t h e s e t o f a l l p o i n t s t h a t s a t i s fy a l l t h e

c o n s t r a i n ts o f t h e l i n e a r p r o g r a m m i n g p r o b l e m is e x a c tl y t h e i n t e r s e c t i o n

o f t h e c l o se d h a l f -s p a c e s d e t e r m i n e d b y t h e c o n s t r a in t s . F r o m t h e p r e v i o u s

re s u l t s w e s e e t h a t t h i s s e t o f p o i n t s , i f i t i s n o n e m p t y , i s a c o n v e x s e t ,

b e c a u s e i t i s t h e i n t e r s e c t i o n o f a f i n it e n u m b e r o f c o n v e x s e ts . I n g e n e ra l ,

t h e i n t e r s e c t i o n o f a f i n it e s e t o f c l o s e d h a l f - s p a c e s i s c a l l e d a

c o n v e x

p o l y h e d ro n , a n d t h u s , i f i t i s n o t e m p t y , t h e s e t o f f e a s i b l e s o l u t i o n s t o a

g e n e r a l l i n e a r p r o g r a m m i n g p r o b l e m i s a c o n v e x p o l y h e d r o n .

W e n o w t u r n t o d e s c r ib i n g t h e p o i n t s a t w h i c h a n o p t i m a l s o l u t i o n t o a

g e n e r a l l i n e a r p r o g r a m m i n g p r o b l e m c a n o c c u r . W e f i r s t m a k e t h e f o l l o w -

ing def in i t ion .

DEFINITION. A po in t

x ~ R i s a c o n v e x c o m b i n a t i o n

o f t h e p o i n t s

X l, X 2 , . . . , x r i n R " i f f o r s o m e r e a l n u m b e r s c 1, c 2 , . . . , c r which sa t i s fy

~ C i - - 1 a n d

C i > 0 , 1 < i < r ,

i = l

w e h a v e

4.

X - - 2 __ , C i X i .

i = 1

THEOREM 1.5.

T h e s e t o f a l l c o n v e x c o m b i n a t i o n s o f a fi n i t e s e t o f p o i n t s

i n R n i s a c o n v e x s e t .

P r o o f Exerc i se . /x

DEFINITION. A po i n t u in a conv ex se t S i s ca l l ed a n ex t rem e po in t o f

S i f i t i s n o t a n i n t e r i o r p o i n t o f a n y l in e s e g m e n t i n S . Th a t is , u i s a n

e x t r e m e p o i n t o f S i f t h e r e a r e n o d i s t i n c t p o i n t s x I a n d x 2 i n S s u c h t h a t

U - - " ,~ X 1 d -

(1

- A ) x 2 , 0

< A 3x + 5y. Thus , we need cons ider only the points in region I to

so lve t he min imiza t ion p r ob lem, s ince i t i s on ly t hose po in t s a t wh ich the

ob jec t ive f unc t ion t akes on va lue s sma l l e r t han 15 . Reg ion I i s c losed and

bo un de d and has a f in i te nu m be r of ext re m e points : (0 ,3 ) , (0 , 1), (2 , 0 ),

( 5 , 0 ) . Consequen t ly , Theor em 1 . 7 app l i e s . By eva lua t ing the ob jec t ive

f unc t ion a t t he se f our po in t s , we f i nd tha t t he min imum va lue i s z = 5 a t

(0 ,1) . No te tha t o th er choices for the d ividing hy perp lane a re poss ib le . ZX

EXAM PLE 6 . Co ns ide r the l i nea r p r og r am m ing p r ob lem

M aximize z = 2x + 3 y

subjec t to

x 3y < 9

2x + 3y < 12

x > _ O , y > _ O .


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90 C h a p t e r 1 I n t r o d u c t io n t o L i n e a r P r o g r a m m i n g

TABLE 1 4

E x t r e m e p o i n t V a l u e o f z = 2 x + 3 y

(0,0) 0

(0,3) 9

(6,0) 12

(3,2) 12

Th e c o n v e x s e t o f a l l f e a s i b l e s o l u t i o n s i s s h o w n i n F i g u re 1 . 1 7 . Th e

e x t r e m e p o i n t s a n d c o r r e s p o n d i n g v a l u es o f th e o b j ec t iv e f u n c t i o n a r e

g i v e n in Ta b l e 1 .4 . W e s e e t h a t b o t h (6, 0 ) a n d (3, 2 ) a r e o p t i m a l s o l u t i o n s

t o t h e p r o b l e m . T h e l i n e s e g m e n t j o i n i n g t h e s e p o i n t s i s

( x , y )

= X (6 , 0 ) + (1 - A ) (3 , 2 )

= ( 6 A , 0 ) + ( 3 - 3 A , 2 - 2 A )

= ( 3 + 3 A , 2 - 2 A ) f o r 0 < A < I .

F o r a n y p o i n t ( x , y ) o n t h i s li n e s e g m e n t w e h a v e

z = 2 x + 3 y = 2 ( 3 + 3 A ) + 3 ( 2 - 2 A )

= 6 + 6A + 6 - 6A

= 12.

A n y p o i n t o n t h is l in e s e g m e n t is a n o p t i m a l s o l u t io n . A

1 4 EXERCISES

In Exercises 1-12 (a) find the extreme points of the set of feasible solutions for

the g iven l inear program ming progr am and (b) f ind the opt imal so lu t ion(s).


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1.4 The Extreme Point Theorem ~1

1. M a x i m i z e z = x + 2 y

s u b j e c t t o

3 x + y < 6

3 x + 4 y < 1 2

x > 0 , y > 0 .

3. M a x i m i z e z = 3 x + y

s u b j e c t t o

- 3 x + y > 6

3x + 5y < 15

x > 0 , y > 0 .

5. M i n i m i z e z = 3 x + 5 y

s u b j e c t t o t h e s a m e c o n s t r a i n t s

a s t h o s e i n E x e r c i s e 4 .

7. M a x i m i ze z = 2 x + 5 y

s u b j e c t t o

2 x + y > 2

x y < 8

x y > 3

2 x + y < 12

x > _ 0 , y > _ 0 .

9 . M a x i m i z e z = 2 x I + 4 x 2 + 3 x 3

s u b j e c t t o

x I + x 2 + x 3 < 12

x I + 3x 2 + 3x 3 < 24

3x I + 6x 2 + 4x 3 < 90

X 1 >_~ 0 , X 2 ~_~ 0 , X 3 >_~ 0 .

1 1. M a x i m i z e z

= 5X 1 +

2 x 2 + 3 x 3

s u b j e c t t o

x I + x 2 + x 3 - 1

2 x I + 5x 2 + 3x 3 < 4

4x I + x 2 + 3x 3 < 2

X 1 >__ 0 , X 2 >__ 0 , X 3 >__ 0 .

1 3. P r o v e T h e o r e m 1 .5 .

1 4. P r o v e T h e o r e m 1 .6 .

2 . M i n i m i z e z = 5 x - 3 y

s u b j e c t t o

x + 2 y < 4

x + 3 y > 6

x > 0 , y > 0 .

4. M a x i m i z e z = 2 x + 3 y

s u b j e c t t o

3 x + y < 6

x + y < 4

x + 2 y < 6

x > _ O , y > 0 .

6 . M a x i m i z e z - - - i x +

s u b j e c t t o

x + 3 y < 6

x + y > 4

x > _ 0 , y > _ 0 .

8 . M a x i m i z e z

= 2 X 1 + 4 X 2

s u b j e c t t o

5X 1 "4- 3X 2 +

5X 3 < 15

1 0 x I + 8 x 2 -1- 1 5 x 3 < 4 0

x 1 >_. 0 , x 2 >_~ 0 , X 3 >_~ 0~

1 0. M i n i m i z e z = 2 X l

+

3x2 + x3

s u b j e c t t o t h e s a m e c o n s t r a i n t s

a s t h o s e i n E x e r c i s e 9 .

1 2 . M i n i m i z e z = 2 X l + x 3

s u b j e c t t o

X 1% - X 2 -~-

X3 -- 1

2 x 1 + x 2 + 2 x 3 >__

3

X 1 >_~ 0 , X 2 ~_~ 0 , X 3 ~ 0 .

1 5. S h o w t h a t a s e t S i n R " i s c o n v e x if a n d o n l y if e v e r y c o n v e x c o m b i n a t i o n o f a

f i n it e n u m b e r o f p o i n t s i n S i s i n S .


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92

C h a p t e r I I n t r o d u c t i o n t o L i n e a r P r o g r a m m i n g

16. Show that if the optimal value of the objective function of a l inear program-

ming problem is attained at several extreme points, then i t is also attained at

any convex combination of these extreme points.

1 5 BA SIC SOLUTIONS

I n t h i s s e c t i o n w e c o n n e c t t h e g e o m e t r i c i d e a s o f S e c t i o n 1 . 3 a n d

S e c t i o n 1 .4 w i t h t h e a l g e b r a i c n o t i o n s d e v e l o p e d i n S e c t i o n 1 .2 . W e h a v e

a l r e a d y s e e n t h e i m p o r t a n t r o l e p l a y e d b y t h e e x t r e m e p o i n t s o f t h e s e t o f

f e a s i b l e s o l u t i o n s i n o b t a i n i n g a n o p t i m a l s o l u t i o n t o a l i n e a r p r o g r a m m i n g

p r o b l e m . H o w e v e r , t h e e x t r e m e p o i n t s a r e d i f f i c u l t t o c o m p u t e g e o m e t r i -

c a ll y w h e n t h e r e a r e m o r e t h a n t h r e e v a r i a b l e s i n th e p r o b l e m . I n th i s

s e c t i o n w e g i v e a n a l g e b ra i c d e s c r i p t i o n o f e x t r e m e p o i n t s t h a t w i l l f a c i l i -

t a t e t h e i r c o m p u t a t i o n . T h i s d e s c r i p t io n u s e s t h e c o n c e p t o f a b a s ic

s o l u ti o n t o a l i n e a r p r o g r a m m i n g p r o b l e m . T o l a y t h e f o u n d a t i o n f o r t h e

d e f i n i t i o n o f a b a s i c s o l u t i o n , w e s h a l l n o w p ro v e t w o v e ry i m p o r t a n t

g e n e r a l t h e o r e m s a b o u t l i n e ar p r o g r a m m i n g p r o b l e m s i n c a n o n i ca l f o r m .

C o n s i d e r t h e l i n e a r p r o g r a m m i n g p r o b l e m i n c a n o n i c a l f o r m

M a x i m i z e z = c Tx (1 )

sub jec t to

Ax = b (2)

x > 0 , ( 3 )

w he re A is an m s m atr ix , c ~ R ~, x ~ R ~, an d b ~ R m. L e t t h e c o l u m n s

o f A b e d e n o t e d b y A 1 , A 2 , . . . , A ~ . W e c a n t h e n w r i t e (2 ) a s

x 1 A 1 + x 2 A 2 + " '" + x s A s = b . (4)

W e m a k e t w o a s s u m p t i o n s a b o u t t h e c o n s t r a i n t m a t r i x A . W e a s s u m e t h a t

m < s a n d t h a t t h e r e a r e m c o l u m n s o f A t h a t a r e l i n e a rl y i n d e p e n d e n t .

T h a t is, t h e r a n k o f A is m . T h e s e a s s u m p t i o n s a r e t r u e f o r a l i n e a r

p r o g r a m m i n g p r o b l e m i n c a n o n i c a l f o r m t h a t a r o s e f r o m a p r o b l e m i n

s t a n d a rd fo rm a s g i v e n i n E q u a t i o n s (4 ) , ( 5 ) , a n d (6 ) i n S e c t i o n 1 . 2 .

T h i s s e t o f m c o l u m n s , a n d i n d e e d a n y s e t o f m l i n e a r l y i n d e p e n d e n t

c o l u m n s o f A , fo rm s a b a s i s fo r

R m.

W e c a n a l w a y s r e n u m b e r t h e c o l u m n s

o f A ( b y r e o r d e r i n g t h e c o m p o n e n t s o f x ), s o t h a t t h e l a s t m c o l u m n s o f A

a re l i n e a r l y i n d e p e n d e n t . L e t S b e t h e c o n v e x s e t o f a ll f e a s i b l e s o l u t i o n s

t o t h e p r o b l e m d e t e r m i n e d b y ( 1) , ( 2) , a n d ( 3) .

THEOREM 1.8. S u p p o s e t h a t t h e la st m c o l u m n s o f A , w h i c h w e d e n o t e b y

t~ 1, t ~ 2 , . . . , t~ m , a r e l i ne a r ly i n d e p e n d e n t a n d s u p p o s e t h a t

Xtl t~ l "Jr- XP2 t~ 2 "Jr- "'" "[- X m t~ m =

b , ( 5 )


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1 . 5 B a s i c S o l u t i o n s 9 3

w h e r e x ' > 0 f o r i = 1, 2 , . . . , m . T h e n th e p o i n t

, ,

, )

X = ( 0 , 0 , . . . , 0 ,

X l ,

X 2 , . . . , x

m

is a n e x t r e m e p o i n t o f S .

P r o o f W e a s s u m e d x > 0 i n t h e s t a t e m e n t o f t h e t h e o r e m . E q u a t i o n

( 5 ) r e p r e s e n t s A x = b , s i nc e t h e f ir st s - m c o m p o n e n t s o f x a r e z e ro .

T h u s , x is a fe a s ib l e s o l u t i o n t o t h e l i n e a r p r o g r a m m i n g p r o b l e m g i v e n b y

( 1 ) , ( 2 ) , a n d ( 3 ) . A s s u m e t h a t x i s n o t a n e x t r e m e p o i n t o f S . T h e n , x l i e s i n

t h e i n t e r i o r o f a l i ne s e g m e n t i n S . T h a t is , t h e r e a r e p o i n t s v a n d w i n S

b o t h d i f f e r e n t f r o m x a n d a n u m b e r A, 0 < A < 1 , s u c h t h a t

N o w

a n d

x = A v + ( 1 - A ) w .

, , , )

u - - ( V l ~ V 2 , . . . ,V s _ m ~ V l , V 2 ~ . . . ~ U m

( 6 )

? ! ! )

W " ~ W 1 , W 2 , . . . , W s _ m , W 1 , W 2 , . . . ~ w m ,

w h e r e a l l th e c o m p o n e n t s o f v a n d w a r e n o n n e g a t i v e , s i n c e v a n d w a r e

f e a s i b l e s o l u t i o n s . S u b s t i t u t i n g t h e e x p r e s s i o n s f o r x , v , a n d w i n t o ( 6) 7 w e

h a v e

0 - -

AU i a -

(1 -

A ) w i ,

1 < i < s - m

(7)

! t

x j = A v ~ + ( 1 - A ) w ) , 1 < j < m .

S i n c e al l t h e t e r m s i n ( 7) a r e n o n n e g a t i v e a n d A a n d 1 - A a r e p o s i t i v e , w e

c o n c l u d e t h a t v i = O a n d w i = 0 f or i = 1 , 2 , . . . , s - r e . S i nc e v is a

f e a s ib l e s o l u t io n , w e k n o w t h a t A v = b a n d , b e c a u s e t h e f ir st s - m

c o m p o n e n t s o f v a r e z e r o , t h is e q u a t i o n c a n b e w r i t t e n a s

l l l

u1A~I -1- u2 Ar 2 "q- "'"

q - U m i ~ m

-"

b . ( 8 )

I f w e n o w s u b t ra c t E q u a t i o n ( 8) f ro m E q u a t i o n ( 5) , w e h a v e

( X q - - V ~ ) / ~ 1 "~ ( X 2 - - v2 ) A t 2

-4- . . . " 4 " ( X t m - v t m ) t ~ m - - O .

S i n c e w e a s s u m e d t h a t

/ ~ l , A r 2 , . . . , A r m w e r e

l i n e a r l y i n d e p e n d e n t , w e

c o n c l u d e t h a t

' ' f o r 1 < i < m

i = V i - - _

a n d c o n s e q u e n t l y t h a t x = v . B u t w e h a d a s s u m e d x # v . T h i s c o n t r a d i c -

t i o n im p l i e s t h a t o u r a s s u m p t i o n t h a t x i s n o t a n e x t r e m e p o i n t o f S i s

f al se . T h u s , x i s a n e x t r e m e p o i n t o f S , a s w e w a n t e d t o s h o w . A

THEOREM 1.9.

I f

x = (X 1,

X 2 , . . . , X , ) is a n e x tr e m e p o i n t o f S , t h e n t h e

c o l u m n s o f A t h a t c o r r e sp o n d to p o s i t i ve x j f o r m a li n e a rl y i n d e p e n d e n t s e t o f

vec tors in R m.


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9 4 C h a p t e r 1 I n t r o d u c ti o n t o L i n e a r P r o g r a m m i n g

P r o o f To s impl if y t he no t a t i on , r en um be r t he co lum ns o f A and the

com pon en t s o f x , i f nece ssa ry , so t ha t t he l a s t k com pon en t s , d eno ted by

! p t

x i , x 2 , . . . , X k, a r e pos i t i ve . Thus , Equa t ion ( 4 ) c an be wr i t t en a s

x~A' 1 + x~A' 2 + "- +x~,A' k = b. (9)

We m us t show tha t A'I, A '2 , . . . , / ~k a r e l i nea r ly i nde pen den t . Sup pose t hey

are l inear ly de pe nd ent . This m ean s tha t the re a re n um be rs c j, 1 _< j _< k ,

not a ll of which a re ze ro , such tha t

C12~ 1 + CEAr2 + ""

+ - Ck 2 ~ k

- - 0 . (10)

Say tha t c t ~ O. Mult ip ly Equa t ion (10) by a pos i t ive sca la r d , and f i r s t add

the r e su l t i ng equa t ion to Equa t ion ( 9 ) , ge t t i ng Equa t ion ( 11) , and second

sub t r ac t i t f r om Equa t ion ( 9 ) , ge t t i ng Equa t ion ( 12) . We now have

( X ~ + d c l ) A ' 1 + ( x ~ + d c 2 ) A r 2

+ " '" + ( X P k + d C k ) t ~ k

= b (11)

( X ~ - - d c l ) A P l + ( x ~ - d c 2 ) A r 2

+ . . . + ( X ~ k - - d C k ) t ~ k

= b . (12)

Now c ons ider the poin ts in R ~,

! ! p

V - - ( 0 , 0 , . . . , 0 , X 1 +

d C l , X 2 + d c 2 , . . . , x k + d c k )

a n d

' - d c ' - d c 2 ' - d c k )

- ( O , O , . . . , O , X l 1 , x 2 , . . . , X k 9

Since d i s any pos i t ive sca la r , we may choose i t so tha t

!

0 < d < m i n xj c j ~ : 0 .

9 I - ~ j l '

1

With th is choice of d , the las t k coordina tes of both v and w are pos i t ive .

Th i s f a c t t oge the r w i th Equa t ions ( 11) and ( 12) impl i e s t ha t v and w a r e

feas ib le solut ions . But we a l so have

1 1

x - ~ v + ~ w ,

con t r ad i c t i ng the hypo thes i s t ha t x i s an ex t r eme po in t o f S . Thus our

a ssum pt ion th a t t he l a st k co lumn s o f A a re l i nea rly de pe nd en t i s f a lse ;

t hey a r e l i nea r ly i ndep ende n t . A

COROLLARY 1.1. I f X is a n e x t r e m e p o i n t a n d X i l , . . . , X i r a r e t h e r

p o s i t i ve c o m p o n e n t s o f x , t h e n r < m , a n d t h e s e t o f c o l u m n s A l l , . . . , A i , c a n

b e e x t e n d e d t o a s e t o f rn l i n ea r ly i n d e p e n d e n t v e c to r s i n R m b y a d j o i n i n g a

s u it a b l y c h o s e n s e t o f r n - r c o l u m n s o f A .

P r o o f .

Exercise . A

THEOREM 1.10. A t m o s t m c o m p o n e n t s o f a n y e x t r e m e p o i n t o f S c a n b e

p o s i t i ve . T h e r e st m u s t b e z e ro .


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1 .5 Bas ic So lu t ions ~

Proo f . T h e o r e m 1 .9 s ay s t h a t t h e c o l u m n s o f A c o r r e s p o n d i n g t o th e

p o s it iv e c o m p o n e n t s o f a n e x t r e m e p o i n t x o f t h e s e t S o f fe a s ib l e

s o l u t i o n s a r e l i n e a r l y i n d e p e n d e n t v e c t o r s i n R m. B u t t h e r e c a n b e n o

m o r e t h a n m l in e a r ly i n d e p e n d e n t v e c to r s i n

R m.

T h e r e f o r e , a t m o s t m o f

t h e c o m p o n e n t s o f x a r e n o n z e r o . A

A n i m p o r t a n t f e a t u r e o f th e c a n o n i c a l f o r m o f a l in e a r p r o g r a m m i n g

p r o b l e m is t h a t t h e c o n s t r a i n t s A x = b f o r m a s y s t e m o f m e q u a t i o n s i n s

u n k n o w n s . O u r a s s u m p t i o n t h a t th e r e a r e m l in e a rl y i n d e p e n d e n t c o l u m n

v e c t o r s m e a n s t h a t t h e r a n k o f A is m , s o th e r o w s o f A a r e a l s o li n e a r l y

i n d e p e n d e n t . T h a t is , r e d u n d a n t e q u a t i o n s d o n o t o c c u r i n t h e s y s t e m o f

c o n s t ra i n ts . I n T h e o r e m s 1 .8 a n d 1 .9 w e s h o w e d t h e r e l a t i o n s h i p s b e t w e e n

t h e e x t r e m e p o i n t s o f t h e s e t S o f f e a s i b l e s o l u t i o n s a n d t h e l i n e a r l y

i n d e p e n d e n t c o l u m n s o f A . W e n o w u s e i n f o r m a t i o n a b o u t s o l u t i o n s to a

s y s t e m o f e q u a t i o n s t o d e s c r i b e f u r t h e r t h e p o i n t s o f S . N o t e t h a t S i s j u s t

t h e s e t o f s o l u t io n s t o A x = b w i t h n o n n e g a t i v e c o m p o n e n t s ( x > 0 ).

C o n s i d e r a s y s te m o f m e q u a t i o n s in s u n k n o w n s ( m < s ) a n d w r i te it

i n m a t r i x f o r m a s A x = b . A s s u m e t h a t a t le a s t o n e s e t o f m c o l u m n s o f A

is l in e a r ly i n d e p e n d e n t . C h o o s i n g a n y s e t T o f m l i n ea r ly i n d e p e n d e n t

c o l u m n s o f A ( w h i c h i s c h o o s i n g a b a s is f o r R m ) , se t t h e s - m v a r i a b l e s

c o r r e s p o n d i n g t o t h e r e m a i n i n g c o l u m n s e q u a l t o z e r o . T h e e q u a t i o n

A x = b m a y b e w r i t t e n a s

x 1 A 1 + x 2 A 2 + " ' " + X s A s =

b , ( 1 3 )

w h e r e A i is t h e i t h c o l u m n o f A . B u t w e h a v e s e t s - m o f t h e v a r i a b l e s

e q u a l t o z e r o . L e t

i l, i 2 , . . . , i m

b e t h e i n d i c e s o f t h e v a r i a b l e s t h a t w e r e

n o t s e t t o z e r o . T h e y a r e a l so t h e i n d i c e s o f t h e c o l u m n s o f A in t h e s e t T .

C o n s e q u e n t l y , ( 1 3 ) r e d u c e s t o

x i l A i l + x i 2 A i 2 + . . . + X i m A i m - - b ,

w h i c h is a s y s te m o f m e q u a t i o n s i n m u n k n o w n s a n d h a s a u n i q u e

s o l u t io n . ( W h y ? ) T h e v a l u e s o f t h e m v a r i a b l e s o b t a i n e d f r o m s o l v in g t h is

s y s t e m a l o n g w i t h t h e s - m z e r o s f o r m a v e c t o r x th a t is c a l l e d a

b a s i c

s o l u t i o n

t o A x

= b .

EX A M PL E 1. C o n s i d e r th e s y s t e m o f t h r e e e q u a t i o n s i n s ix u n k n o w n s

x 1

[ I 1

0 1 0 1 0

x3

0 - 1 - 1 0 - 1 - 1 - b2 .

1 2 2 1 1 1 x4 b3

x 5

x 6

- -


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~ Cha pt e r 1 In t rod uc t i on t o L i ne ar Programm i ng

S e t t i n g x 2 - - X 3 - - X 5 = 0 , w e g e t t h e s y s t e m o f t h r e e e q u a t i o n s i n t h r e e

u n k n o w n s g i v e n b y

[1 0 0 ]I X l i b l l

0 - 1 x 4 - - b2 9

1 1 1

x 6 b 3

T h e c o l u m n s o f t h is c o e ff i ci e n t m a t r i x a r e l i n e a r ly i n d e p e n d e n t , a n d t h i s

sys tem ha s the so l u t i on x I = b~ , x 4 = b 2 + b 3 - b l , x 6 - - b 2 . C on se -

q u e n t l y , a b a s i c s o l u t i o n t o t h e o r i g i n a l s y s t e m i s

X = ( b l , 0 , 0 , b 2 + b 3 - b l , 0 , - b 2 ) .

O n t h e o t h e r h a n d , i f w e s e t x 1

- - X 3 = X 5 " -- 0 , w e obtain the system

[o o o ]I x 2 1 b l l

1 0 - 1 x 4 -- b2 9

2 1 1 x 6 b 3

H e r e t h e c o l u m n s o f t h e c o e f fi c ie n t m a t r i x a r e n o t l in e a r ly i n d e p e n d e n t . I n

f a c t, c o l u m n 1 i s t h e s u m o f c o l u m n s 2 a n d 3 . T h i s s y s t e m c a n n o t b e s o l v e d

i f b I ~: 0 . Co n s e q u e n t ly , t h i s c h o i c e o f v a r i a b l e s d o e s n o t l e a d t o a b a s i c

s o lu t i o n . / x

I n a n y b a s i c so l u t i o n , t h e s - m v a r i a b l e s t h a t a r e s e t e q u a l t o z e r o a r e

c a l l e d n o n b a s i c v a r i a b l e s , a n d t h e m v a r i a b l e s s o l v e d f o r a r e c a l l e d b a s i c

v a r i a b l e s . A l t h o u g h t h e t e r m

b a s i c s o l u t i o n

a p p e a r s i n a l l t h e l i t e r a t u r e

d e s c r i b i n g l i n e a r p r o g r a m m i n g , i t c a n b e m i s l e a d i n g . A b a s i c s o l u t i o n i s a

s o lu t i o n t o t h e s y s t e m A x = b; i t d o e s n o t n e c e s s a r i l y s a t i s fy x >_ 0 , a n d

t h e r e f o r e i t i s n o t n e c e s s a r i l y a f e a s i b l e s o l u t i o n t o t h e l i n e a r p r o g r a m m i n g

p r o b l e m g iv e n b y ( 1 ) , ( 2 ) , a n d ( 3 ) .

D EFINIT IO N. A b a s i c fe a s i b l e s o l u t i o n t o t h e l i n e a r p r o g r a m m i n g p r o b -

l e m g iv e n b y ( 1 ) , ( 2 ) , a n d ( 3 ) i s a b a s i c s o lu t i o n t h a t i s a l s o f e a s ib l e .

THEOREM 1.11. F o r t h e l in e a r p r o g r a m m i n g p r o b l e m d e t e r m i n e d b y (1) ,

(2) , a n d (3), ever y b a si c f ea s ib l e s o l u t i o n i s a n ex t r em e p o i n t , a n d , co n v er s e ly ,

every ex t rem e p o in t i s a bas i c f eas ib l e so lu t ion .

Proof . E x e r c i s e . A

THEOREM 1.12. T h e p r o b l e m d e t e r m i n e d b y (1) , (2) , a n d (3) h a s a f i n i t e

n u m b e r o f b a s i c fe a s i b le s o lu t io n s .

P r o o f T h e n u m b e r o f b a s ic s o l u ti o n s t o th e p r o b l e m is

( s ) s, ( ) s

m m ! ( s - m ) ! s - m


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1.5 Ba sic Solut ion s ~

b e c a u s e s - m v a r i a b l e s m u s t b e c h o s e n t o b e s e t t o ze r o , o u t o f a t o t a l o f

s v a r i ab l e s . T h e n u m b e r o f b a s i c f e a s ib l e s o l u t i o n s m a y b e s m a l l e r t h a n

t h e n u m b e r o f b a si c s o l u ti o n s , s i n ce n o t a l l b a s i c s o lu t i o n s n e e d t o b e

feas ib le . A

W e n o w e x a m i n e t h e r e l a t i o n s h i p b e t w e e n t h e s e t S o f f e a s ib l e s o l u-

t i o n s t o a s t a n d a r d l i n e a r p r o g r a m m i n g p r o b l e m g i v e n i n E q u a t i o n s ( 4 ) ,

( 5 ), a nd (6 ) in Se c t ion 1 .2 a nd the se t S ' o f f e a s ib le so lu t ion s to th e

a s s o c i a t e d c a n o n i c a l l i n e a r p r o g r a m m i n g p r o b l e m g i v e n i n E q u a t i o n s ( 1 2 ) ,

( 13 ) , a nd (14 ) in Se c t ion 1 .2 . W e ha ve a l r e a dy d i sc usse d the me thod o f

a dd ing s l a c k va r i a b le s to go f rom a po in t in S to a po in t in S ' . C o nve r se ly ,

w e t r u n c a t e d v a r i a b l e s t o m o v e f r o m S ' t o S . M o r e s p e c i f i c a l l y , w e h a v e

t h e f o ll o w i ng t h e o r e m , w h o s e p r o o f w e l e av e a s a n e x e rc i se .

THEOREM 1.13. E v e ry e x t r em e p o i n t o f S yi e ld s a n e x tr e m e p o i n t o f S '

wh en s lack var iables are add ed . Co nverse ly , every ex t reme po in t o f S ' , wh en

t runc a t e d , y ie l ds an e x t r e m e po i n t o f S .

P r o o f Exe rc i se . A

THEOREM 1.14.

The con vex se t S o f a l l f eas ib le so lu t ions to a l inear

p r o g r a m m i n g p r o b l e m i n st a n d a r d f o r m h a s a f in i t e n u m b e r o f e x tr e m e p o i n t s .

P r o o f

Exerc ise . /x

B y c o m b i n i n g th e E x t r e m e P o i n t T h e o r e m ( T h e o r e m 1 .7) a n d T h e o r e m

1 .1 4 w e c a n g i v e a p r o c e d u r e f o r s o lv i ng a s t a n d a r d l i n e a r p r o g r a m m i n g

prob le m g ive n by Equa t ions (4 ) , ( 5 ) , a nd (6 ) in Se c t ion 1 .2 . F i r s t , s e t up the

a s s o c i a te d c a n o n i c a l f o r m o f t h e p r o b l e m . T h e n f in d a ll b a s i c s o l u ti o n s a n d

e l i m i n a t e t h o s e t h a t a r e n o t f e a s i b l e . F i n d t h o s e t h a t a r e o p t i m a l a m o n g

the ba s i c f e a s ib le so lu t ions . S inc e the ob je c t ive func t ion doe s no t c ha nge

b e t w e e n t h e s t a n d a r d a n d c a n o n i c a l f o rm s o f t h e p r o b l e m , a n y o p ti m a l

so lu t ion to the c a non ic a l f o rm , found a s de sc r ibe d a bove , i s a n op t ima l

s o l u t i o n t o t h e s t a n d a r d p r o b l e m .

F r o m t h e s i t u a ti o n u n d e r c o n s i d e r a t i o n , t h e n u m b e r o f b a s ic s o lu t i o n s

t o b e e x a m i n e d i s n o m o r e t h a n

m + n ) .

n

Al though th i s numbe r i s f in i t e , i t i s s t i l l t oo l a rge fo r a c tua l p rob le ms . For

e x a m p l e , a m o d e r a t e - s i z e p r o b l e m w i t h m = 2 0 0 a n d n = 3 00 w o u l d h a v e

a b o u t

10144

basic so lu t ions .

EX AM PLE 2 . C o n s i d e r t h e l i n e a r p r o g r a m m i n g p r o b l e m g i v e n i n E x a m -

p le 3 in Se c t ion 1 .2 . I n th i s e xa mple we c a n se l e c t two o f the four va r i a b le s

x , y , u , v a s n o n b a s i c v a r i a b l e s b y s e t t i n g t h e m e q u a l t o z e r o a n d t h e n


29/32

98

C h a p t e r 1 I n t r o d u c t io n to L i n e a r P r o g r a m m i n g

s o l v e f o r t h e b a s i c v a r i a b l e s . I f i n

w e s e t

t h e n

2 2 1 0

5 3 0 1

x

u

u

8

x = y = 0 ( n o n b a s i c v a r i a b l e s ) ,

u = 8 a n d v = 1 5 ( b a s i c v a r i a b l e s ) .

T h e v e c t o r [0 0 8 1 5] T i s a b a s i c f e a s ib l e s o l u t i o n t o t h e c a n o n i c a l

f o r m o f t h e p r o b l e m a n d h e n c e a n e x t r e m e p o i n t o f S ' . B y t r u n c a t i n g t h e

s l a c k v a r i a b l e s w e g e t [ 0 0] T, w h i c h i s a n e x t r e m e p o i n t o f S a n d a

f e a si b le s o l u t i o n t o t h e s t a n d a r d f o r m o f th e p r o b l e m .

I f i n s t e a d w e s e t

t h e n

T h e v e c t o r [ 0

u i s n e g a t i v e .

x = v = 0 ( n o n b a s i c v a r i a b l e s ) ,

y = 5 a n d u = - 2 ( b as ic v a r i a b l es ) .

5 - 2

0 ] T i s a b a s i c s o lu t i o n , b u t i t is n o t f e a s ib l e , s i n c e

I n T a b l e 1 .5 w e l i st al l t h e b a s i c s o l u t i o n s , i n d i c a t e w h e t h e r t h e y a r e

f e a si b le , a n d g i ve t h e t r u n c a t e d v e c t o r s . T h e s t u d e n t s h o u l d l o c a t e t h e s e

t r u n c a t e d v e c t o r s o n F i g u r e 1 . 1 4 . O n c e w e d i s c a r d t h e b a s i c s o l u t i o n s t h a t

a r e n o t f e a s i b l e , w e s e l e c t a b a s i c f e a s i b l e s o l u t i o n f o r w h i c h t h e o b j e c t i v e

f u n c t i o n is a m a x i m u m . T h u s , w e a g a in o b t a i n t h e o p t i m a l s o l u t i o n

3 5

x

= ~ , J v = ~ , u = v , v = v

TABLE 1.5

V a l u e o f T r u n c a t e d

x y u v T y p e o f s o l u t i o n z = 1 2 0 x + 1 0 0 y v e c t o r

0 0 8 15 Bas ic f eas ib le 0 (0 , 0 )

0 4 0 3 Bas ic f eas ib le 400 (0 , 4 )

0 5 - 2 0 Bas ic , no t f ea s ib le m (0 , 5 )

4 0 0 - 5 Bas ic , no t f ea s ib l e m (4 , 0 )

3 0 2 0 Bas ic f eas ib le 360 (3 , 0 )

3 5

~ 0 0 Bas ic f eas ib le 430 (3 , 5 )


30/32

1.5 Basic Solutions

99

1 5 EXERCISES

1 . Suppose the canon ica l fo rm of a l ine r p rogramming p rob lem i s g iven by the

cons t ra in t ma t r ix A and re source vec to r b , where

A - 2 1 0 0 0 a n d b = 3 .

4 0 3 0 1 6

De te rmine which o f the fo l lowing po in t s i s

( i ) a feas ib le so lu t ion to the l inea r p rogramming p rob lem.

( i i ) an extreme point of the se t of feas ible solut ions .

(i i i) a basic solution.

(iv) a basic feasible solution.

For each bas ic feas ible solut ion x given below, l i s t the bas ic var iables .

3 1 1

0 0 ~ ~ 1

3 3 0 1 1

(a) 0 (b) 5 (c) 0 (d) 1 (e) ~

1 3

5 0 ~ 0

1

6 - 9 0 2

In Exerc i se s 2 and 3 , s e t up a l inea r p rogramming mode l fo r the s i tua t ion

descr ibed. Sketch the se t of feas ible solut ions and f ind an opt imal solut ion by

examin ing the ex t reme po in t s .

2 . The Savory Po ta to Chip C om pany ma kes p izza - f lavored and ch i l i -f l avored

po ta to ch ips . These ch ips mus t go th rough th ree ma in p roces ses : f ry ing ,

f lavoring, and packing. Each ki logram of pizza-f lavored chips takes 3 min to fry ,

5 min to f lavor , and 2 min to pack. Each ki logram of chi l i - f lavored chips takes 3

min to f ry , 4 min to f l avor , and 3 min to pack . The ne t p ro f i t on each k i logram

of pizza chips is $0.12, whereas the ne t prof i t on each ki logram of chi l i chips is

$0.10. The fryer is avai lable 4 hr each day, the f lavorer is avai lable 8 hr each

day, and the packer is avai lable 6 hr each day. Maximize the ne t prof i t wi th

your mode l .

3 . Sugary Donut Bakers , Inc . , i s known for i t s excel lent g lazed doughnuts . The

f i rm a l so bakes doughnuts , which a re then d ipped in powdered suga r . I t makes

a p ro f it o f $0 .07 pe r g lazed doughnu t and $0.05 pe r pow dered suga r dough nut .

The th ree ma in ope ra t ions a re bak ing , d ipp ing ( fo r the powdered suga r

doughnuts on ly ) , and g laz ing ( fo r the g lazed doughnuts on ly ) . Each day the

plant has the capaci ty to bake 1400 doughnuts , d ip 1200 doughnuts , and glaze

1000 doughnuts . Th e m ana ger has ins t ruc ted tha t a t l ea s t 600 g lazed dough-

nu t s mus t be made each day . Maximize the to ta l p ro f i t wi th your mode l .

4 . For Exerc i se 2 , wr i t e the l inea r p rogra m min g p rob lem in canon ica l fo rm,

compute the va lues o f the s l ack va r iab le s fo r an op t ima l so lu t ion , and g ive a

phys ica l in terpre ta t ion for these va lues . Also ident i fy the bas ic var iables of the

opt imal solut ion.


31/32

1 O0 Cha pter 1 In troduct ion to Linear Programm ing

5 . For Exerc i se 3 , wr i t e the l inea r p rogra mm ing p ro b lem in canon ica l fo rm,

compute the va lues of the s lack variables for an opt imal solut ion, and give a

phys ica l in terpre ta t ion for these va lues . Also ident i fy the bas ic var iables of the

opt imal solut ion.

6 . Cons ider the sys tem of equat ions Ax = b , where

A = [ 2 3 4 0 4 ] a nd b = [ 2 ]

1 0 0 -2 1 0

Determine whether each of the fol lowing 5- tuples is a bas ic solut ion to the

system.

(a) (1, 0, 1, 0, 0) (b) (0, 2, - 1, 0, 0)

(c) (2, - 2, 3, 0, - 2) (d) (0, 0, x

, 0 , 0 )

7. Cons id er the sys tem of equa t ions Ax - b , wh ere

[ 2 3 1 0 0 ] [ 1 ]

A - - 1 1 0 2 1 a n d b = 1 .

0 6 1 0 3 4

Determine which of the fol lowing 5- tuples are bas ic solut ions to the sys tem.

Give reasons .

(a) (1 , 0 , - 1 , 1 , 0 )

( b ) ( 0 , 2 , - 5 , 0 , - 1 )

(c) (0, 0, 1, 0, 1)

8 . Cons ide r the l inea r p rogramming p rob lem

M a x i m i z e z = 3 x + 2 y

subjec t to

2 x - y < 6

2x + y < 10

x > _ 0 , y > _ 0 .

(a ) T rans fo rm th i s p rob lem to a p rob lem in canon ica l fo rm.

(b) For each extreme point of the new problem, ident i fy the bas ic var iables .

(c) Solve the problem geometr ica l ly .

9. Con s ide r the l inear p rogram min g p rob lem

Maximize z

= 4x 1 + 2 X 2 + 7X 3

subjec t to

2x 1 - x 2 + 4x 3_< 18

4x I + 2x 2

+ 5 X 3 ___~ 0 , X 2 >_~ 0 , X 3 >_~ 0 .

(a ) T rans fo rm th i s p rob lem to a p rob lem in canon ica l fo rm.

(b ) For each ex t reme po in t o f the new pro b lem, iden ti fy the bas ic va r iab les .

(c ) Which o f the ex t reme po in t s a re op t ima l so lu t ions to the p rob lem?


32/32

1.5 Ba sic Solut ions 101

1 0 . C o n s i d e r t h e l i n e a r p r o g r a m m i n g i n s t a n d a r d f o r m

M a x i m i z e z = c T x

s u b j e c t t o

A x < b

x > 0 .

S h o w t h a t t h e c o n s t r a i n t s A x < b m a y b e w r i t t e n a s

[ x ]

( i) [A I I ] x ' = b

o r a s

( ii ) A x + I x ' = b ,

w h e r e x ' i s a v e c t o r o f s l a c k v a r i a b l e s .

1 1 . P r o v e C o r o l l a r y 1 . 1 ( H i n t : U s e T h e o r e m 0 . 1 3 . )

1 2. P r o v e f o r a l i n e a r p r o g r a m m i n g p r o b l e m i n c a n o n i c a l f o r m t h a t a p o i n t i n t h e

s e t o f f e a s ib l e s o l u t i o n s is a n e x t r e m e p o i n t i f a n d o n l y i f i t i s a b a s ic f e a s i b l e

s o l u t i o n ( T h e o r e m 1 . 1 1 ) .

( H i n t :

U s e T h e o r e m 0 . 1 3 . )

1 3. P r o v e t h a t t h e s e t o f f e a si b l e s o l u t i o n s t o a li n e a r p r o g r a m m i n g p r o b l e m i n

s t a n d a r d f o r m h a s a f in i te n u m b e r o f e x t r e m e p o i n t s ( T h e o r e m 1 .1 4).

Further Reading

Chvfital, V a~ek. Linear Programm ing . Freem an, New Y ork, 1980.

Griinbaum, B.

Co nvex Po ly topes .

Wiley-Interscience, New York, 1967.

Hadley, G. Linear A lgebra . Addison-Wesley, Reading, MA, 1961.

Murty, Katta G. Linear P rogramming . Wiley, New York, 1983.

Taha , Ham dy A. Operat ions Research: An In troduct ion , third ed., Macmillan, N ew York, 1982.

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